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Calculus 2 : Polar

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Example Questions

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Example Question #671 : Calculus Ii

Rewrite the polar equation

$r = \sin ^{2} \theta$

in rectangular form.

Possible Answers:

$y ^{4} = \left (x^{2} +y^{2} \right )^{2}$

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

$y ^{3} = \left (x^{2} +y^{2} \right )^{2}$

$y ^{4} = \left (x^{3} +y^{3} \right )^{2}$

$y ^{3} = \left (x^{2} +y^{2} \right )^{4}$

Correct answer:

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

Explanation:

$r = \sin ^{2} \theta$

$r^{2} \cdot r = r^{2} \sin ^{2} \theta$

$r^{3} =\left ( r \sin \theta \right ) ^{2}$

$\left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}$

$\left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{3} = y ^{4}$

or $y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

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Example Question #1 : Polar

Rewrite in polar form:

$x^{2} = 3xy + 1$

Possible Answers:

$r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$

$r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Correct answer:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Explanation:

$x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

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Example Question #1 : Polar

Rewrite the polar equation

$r ^{2} + 1 = \cos \theta$

in rectangular form.

Possible Answers:

$\sqrt{x^{2} + y ^{2}} =\frac{ y}{x^{2} + y ^{2}}$

$\sqrt{x^{2} + y ^{2}} =\frac{x+ y}{{x^{2} + y ^{2}} +1}$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} }$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

$\sqrt{x^{2} + y ^{2}} =\frac{ y}{{x^{2} + y ^{2}} +1}$

Correct answer:

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

Explanation:

$r ^{2} + 1 = \cos \theta$

$r\left ( r ^{2} + 1 \right )= r \cos \theta$

$\sqrt{x^{2} + y ^{2}} \left ( {x^{2} + y ^{2}} +1 \right ) = x$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

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Example Question #1 : Polar

Give the polar form of the equation of the line with intercepts (0,4), (-6,0) .

Possible Answers:

$r = \frac{12}{3 \sin \theta + 2 \cos \theta }$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

$r = 12\tan \theta$

$r = \frac{12}{ \tan \theta }$

$r = \frac{12}{2 \sin \theta - 3 \cos \theta }$

Correct answer:

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

Explanation:

This line has slope $\frac{4-0}{0- (-6)} = \frac{2}{3}$ and -intercept (0.4) , so its Cartesian equation is $y = \frac{2}{3}x + 4$ .

By substituting, we can rewrite this:

$r \sin \theta = \frac{2}{3}r \cos \theta + 4$

$r \sin \theta - \frac{2}{3}r \cos \theta = 4$

$3 r \sin \theta - 2 r \cos \theta = 12$

$r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

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Example Question #1 : Polar

Give the rectangular coordinates of the point with polar coordinates

$\left ( -4 , \frac{\pi }{3}\right )$ .

Possible Answers:

$\left ( -2 \sqrt{3} , -2\right )$

$\left ( 2, -2 \sqrt{3} \right )$

$\left ( -2 \sqrt{3} , 2\right )$

$\left ( -2, -2 \sqrt{3} \right )$

$\left ( 2 \sqrt{3} , -2\right )$

Correct answer:

$\left ( -2, -2 \sqrt{3} \right )$

Explanation:

$x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$

$y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}$

The point will have rectangular coordinates $\left (- 2, -2 \sqrt{3} \right )$ .

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Example Question #1 : Polar Form

What would be the equation of the parabola $\small y=x^{2}$ in polar form?

Possible Answers:

$\small r=tan\theta sec\theta$

$\small \small \small r=tan\theta sin\theta$

$\small \small r=cot\theta csc\theta$

$\small r=cot\theta cos\theta$

$\small r=\theta ^{2}$

Correct answer:

$\small r=tan\theta sec\theta$

Explanation:

We know $\small y=rsin\theta$ and $\small x=rcos\theta$ .

Subbing that in to the equation $\small y=x^{2}$ will give us $\small rsin\theta=r^{2}cos^{2}\theta$ .

Multiplying both sides by $\small 1/(rcos^{2}\theta)$ gives us

$\small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta$ .

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Example Question #2 : Polar Form

A point in polar form is given as $\left(3,\frac{\pi}{3}\right)$ .

Find its corresponding (x,y) coordinate.

Possible Answers:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

$\left(\frac{3\sqrt3}{2},\frac{3}{2}\right)$

$\left(3,\frac{3\sqrt3}{2}\right)$

(0,3)

$\left(\frac{3}{2},\frac{3}{2}\right)$

Correct answer:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

Explanation:

To go from polar form to cartesion coordinates, use the following two relations.

$x=rcos(\theta)$

$y=rsin(\theta)$

In this case, our is and our $\theta$ is $\frac{\pi}{3}$ .

Plugging those into our relations we get

$x=\frac{3}{2}$

$y=\frac{3\sqrt3}{2}$ ,

which gives us our (x,y) coordinate.

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Example Question #3 : Polar Form

What is the magnitude and angle (in radians) of the following cartesian coordinate?

(3,4)

Give the answer in the format below.

$(r,\theta)$

Possible Answers:

(5,0.927)

(3,1.04)

(4,0.927)

(12,0.707)

(5,0.643)

Correct answer:

(5,0.927)

Explanation:

Although not explicitly stated, the problem is asking for the polar coordinates of the point (3,4) . To calculate the magnitude, , calculate the following:

$r=\sqrt{x^2+y^2}$

r=5

To calculate $\theta$ , do the following:

$\theta=tan^{-1}\left(\frac{y}{x}\right)$ in radians. (The problem asks for radians)

$\theta = 0.927$

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Example Question #2 : Polar Form

What is the following coordinate in polar form?

(-2,-5)

Provide the angle in degrees.

Possible Answers:

(10,248)

$(\sqrt{29},68)$

(16,112)

$(\sqrt{29},202)$

$(\sqrt{29},248)$

Correct answer:

$(\sqrt{29},248)$

Explanation:

To calculate the polar coordinate, use

$r=\sqrt{x^2+y^2}$

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

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Example Question #201 : Ap Calculus Bc

What is the equation $y=2x^{2}$ in polar form?

Possible Answers:

$r=\frac{1}{2}\sin \theta}{\cos \theta$

$r=\frac{1}{2}\cos \theta}{\tan \theta$

$r=\frac{1}{2}\sin \theta}{\sec \theta$

$r=\frac{1}{2}\tan \theta}{\sec \theta$

$r=\frac{1}{2}\sin \theta}{\tan \theta$

Correct answer:

$r=\frac{1}{2}\tan \theta}{\sec \theta$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=2x^{2}$ , then $r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ .

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

$\frac{\tan \theta}{\cos \theta}=2r$

$\tan \theta}{\sec \theta=2r$

$\frac{1}{2}\tan \theta}{\sec \theta=r$

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Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #671 : Calculus Ii

Example Question #1 : Polar

Example Question #1 : Polar

Example Question #1 : Polar

Example Question #1 : Polar

Example Question #1 : Polar Form

Example Question #2 : Polar Form

Example Question #3 : Polar Form

Example Question #2 : Polar Form

Example Question #201 : Ap Calculus Bc

All Calculus 2 Resources

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